Solve (sqrt{3}x-2)^2+x^2=2^2

$$( \sqrt { 3 } x - 2 ) ^ { 2 } + x ^ { 2 } = 2 ^ { 2 }$$

Solve for x

$$x=\sqrt{3}\approx 1.732050808$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}x-2\right)^{2}.

$$\left(\sqrt{3}\right)^{2}x^{2}-4\sqrt{3}x+4+x^{2}=2^{2}$$

The square of \sqrt{3} is 3.

$$3x^{2}-4\sqrt{3}x+4+x^{2}=2^{2}$$

Combine 3x^{2} and x^{2} to get 4x^{2}.

$$4x^{2}-4\sqrt{3}x+4=2^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-4\sqrt{3}x+4=4$$

Subtract 4 from both sides.

$$4x^{2}-4\sqrt{3}x+4-4=0$$

Subtract 4 from 4 to get 0.

$$4x^{2}-4\sqrt{3}x=0$$

Factor out x.

$$x\left(4x-4\sqrt{3}\right)=0$$

To find equation solutions, solve x=0 and 4x-4\sqrt{3}=0.

$$x=0 x=\sqrt{3}$$